JEE MAIN - Mathematics Hindi (2002 - No. 8)
यदि $$f\left( 1 \right) = 1,{f^1}\left( 1 \right) = 2,$$ तो
$$\mathop {\lim }\limits_{x \to 1} {{\sqrt {f\left( x \right)} - 1} \over {\sqrt x - 1}}$$ है
$$\mathop {\lim }\limits_{x \to 1} {{\sqrt {f\left( x \right)} - 1} \over {\sqrt x - 1}}$$ है
$$2$$
$$4$$
$$1$$
$${1 \over 2}$$